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d , ( x : 7 + 15 ) . 23 + 391 => Đề thiếu
e , ( 19 . x + 2 . 52 ) : 14 = ( 13 - 8 ) 2 - 42
=> ( 19 . x + 2 . 52 ) : 14 = 52 - 16
=> ( 19 . x + 2 . 52 ) : 14 = 25 - 16 = 9
=> 19 . x + 2 . 52 = 9 x 14 = 126
=> 19 . x + 2 . 25 = 126
=> 19 . x + 50 = 126
=> 19 . x = 126 - 50 = 76
=> x = 76 : 19 = 4
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
a) \(\Rightarrow5\left(x-10\right)=10\)
\(\Rightarrow x-10=2\Rightarrow x=12\)
b) \(\Rightarrow3\left(70-x\right)+5=92\)
\(\Rightarrow3\left(70-x\right)=87\)
\(\Rightarrow70-x=29\Rightarrow x=41\)
c) \(\Rightarrow230+\left[16+\left(x-5\right)\right]=315\)
\(\Rightarrow11+x=85\Rightarrow x=74\)
d) \(\Rightarrow707:\left(2^x-5+74\right)=7\)
\(\Rightarrow2^x-5+74=101\Rightarrow2^x-5=27\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
a
\(230+\left[16+\left(y-5\right)\right]=315\cdot1\)
\(230+\left[16+\left(y-5\right)\right]=315\)
\(16+\left(y-5\right)=315-230\)
\(16+\left(y-5\right)=85\)
\(y-5=85-16\)
\(y-5=69\)
\(y=69+5\)
\(y=74\)
b
\(707:\left[\left(2^y-5\right)+74\right]=16-9\)
\(707:\left[\left(2^y-5\right)+74\right]=7\)
\(\left(2^y-5\right)+74=707:7\)
\(\left(2^y-5\right)+74=101\)
\(2^y-5=101-74\)
\(2^y-5=27\)
\(2^y=27+5\)
\(2^y=32\)
\(2^y=2^5\)
\(\Rightarrow y=5\)
1.
a.
11 + x : 5 = 13
x : 5 = 13 - 11
x : 5 = 2
x = 2 . 5
x = 10
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
a, 128 - 3.( x + 4 ) = 23 b, [( 6x - 39 ) : 7 ] . 4 = 12 c, ( x : 3 - 4 ) . 5 = 15
3. ( x + 4 ) = 128 - 23 [(6x - 39 ) : 7 ] = 3 ( x : 3 - 4 ) = 3
3. ( x + 4 ) = 105 ( 6x - 39 ) = 21 x : 3 = 7
x + 4 = 35 6x = 21 + 39 x = 7 . 3
x = 35 - 4 6x = 60 x = 21
x = 31 x = 10 Vậy x = 21
Vậy x = 31 Vậy x = 10
Trl:
a. 128 - 3 . ( x + 4 ) = 23
=> 3 . ( x + 4 ) = 128 - 23
=> 3 . ( x + 4 ) = 105
=> x + 4 = 105 : 3
=> x + 4 = 35
=> x = 35 - 4
=> x = 31
Vậy x = 31
b. [( 6x - 39 ) : 7 ] . 4 = 12
=> [( 6x - 39 ) : 7 ] = 12 : 4
=> [( 6x - 39 ) : 7 ] = 3
=> ( 6x - 39 ) = 3 . 7
=> ( 6x - 39 ) = 21
=> 6x = 21 + 39
=> 6x = 60
=> x = 60 : 6
=> x = 10
Vậy x = 10
c. ( x : 3 - 4 ) . 5 = 15
=> ( x : 3 - 4 ) = 15 : 5
=> ( x :3 - 4 ) = 3
=> x : 3 = 3 + 4
=> x : 3 = 7
=> x = 7 . 3
=> x = 21
Vậy x = 21
d. | x + 2 | = 0
=> x + 2 = 0
=> x = 0 - 2
=> x = -2
Vậy x = -2
e. | x - 5 | = |-7|
=> | x - 5 | = 7
=> x - 5 = 7 hoặc x - 5 = -7
=> x = 7 + 5 hoặc x = -7 + 5
=> x = 12 hoặc x = -2
Vậy \(x\in\left\{12;-2\right\}\)
a) x = 74
b) x = 5
c) x = 3
d) x = 14